Problem: Is ${462429}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {462429}= &&{4}\cdot100000+ \\&&{6}\cdot10000+ \\&&{2}\cdot1000+ \\&&{4}\cdot100+ \\&&{2}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {462429}= &&{4}(99999+1)+ \\&&{6}(9999+1)+ \\&&{2}(999+1)+ \\&&{4}(99+1)+ \\&&{2}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {462429}= &&\gray{4\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {4}+{6}+{2}+{4}+{2}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${462429}$ is divisible by $9$ if ${ 4}+{6}+{2}+{4}+{2}+{9}$ is divisible by $9$ Add the digits of ${462429}$ $ {4}+{6}+{2}+{4}+{2}+{9} = {27} $ If ${27}$ is divisible by $9$ , then ${462429}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${462429}$ must also be divisible by $9$.